Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $t = \dfrac{z^2 - 5z - 36}{-8z^2 + 72z} \div \dfrac{z + 4}{z + 6} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $t = \dfrac{z^2 - 5z - 36}{-8z^2 + 72z} \times \dfrac{z + 6}{z + 4} $ First factor the quadratic. $t = \dfrac{(z + 4)(z - 9)}{-8z^2 + 72z} \times \dfrac{z + 6}{z + 4} $ Then factor out any other terms. $t = \dfrac{(z + 4)(z - 9)}{-8z(z - 9)} \times \dfrac{z + 6}{z + 4} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac{ (z + 4)(z - 9) \times (z + 6) } { -8z(z - 9) \times (z + 4) } $ $t = \dfrac{ (z + 4)(z - 9)(z + 6)}{ -8z(z - 9)(z + 4)} $ Notice that $(z - 9)$ and $(z + 4)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac{ \cancel{(z + 4)}(z - 9)(z + 6)}{ -8z(z - 9)\cancel{(z + 4)}} $ We are dividing by $z + 4$ , so $z + 4 \neq 0$ Therefore, $z \neq -4$ $t = \dfrac{ \cancel{(z + 4)}\cancel{(z - 9)}(z + 6)}{ -8z\cancel{(z - 9)}\cancel{(z + 4)}} $ We are dividing by $z - 9$ , so $z - 9 \neq 0$ Therefore, $z \neq 9$ $t = \dfrac{z + 6}{-8z} $ $t = \dfrac{-(z + 6)}{8z} ; \space z \neq -4 ; \space z \neq 9 $